Problem: Simplify and expand the following expression: $ \dfrac{r}{r - 5}-\dfrac{5r}{4r - 6} $
Solution: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(r - 5)(4r - 6)$ Multiply the first term by $\dfrac{4r - 6}{4r - 6}$ $ \begin{align*} \dfrac{r}{r - 5} \times \dfrac{4r - 6}{4r - 6} & = \dfrac{(r)(4r - 6)}{(r - 5)(4r - 6)} \\ & = \dfrac{4r^2 - 6r}{(r - 5)(4r - 6)}\end{align*} $ Multiply the second term by $\dfrac{r - 5}{r - 5}$ $ \begin{align*} \dfrac{5r}{4r - 6} \times \dfrac{r - 5}{r - 5} & = \dfrac{(5r)(r - 5)}{(4r - 6)(r - 5)} \\ & = \dfrac{5r^2 - 25r}{(4r - 6)(r - 5)}\end{align*} $ Now we have: $ = \dfrac{4r^2 - 6r}{(r - 5)(4r - 6)} - \dfrac{5r^2 - 25r}{(4r - 6)(r - 5)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{4r^2 - 6r - (5r^2 - 25r)}{(r - 5)(4r - 6)} $ $ = \dfrac{4r^2 - 6r - 5r^2 + 25r}{(r - 5)(4r - 6)} $ $ = \dfrac{-r^2 + 19r}{(r - 5)(4r - 6)}$ Expand the denominator: $ = \dfrac{-r^2 + 19r}{4r^2 - 26r + 30}$